[leetcode] 2. Pascal's Triangle II

我是按难度往下刷的，第二道是帕斯卡三角形二。简单易懂，题目如下：

Given an index k, return the k^th row of the Pascal‘s triangle.

For example, given k = 3, Return [1,3,3,1].

Note: Could you optimize your algorithm to use only O(k) extra space?

就是输入k然后输出第k行帕斯卡三角【其实我一直把它叫杨辉三角】。

我这边思路是拿queue来不断弹出压入处理做那个二项式展开：

vector<int> getRow(int rowIndex)
{
    int aspros, defteros;
    aspros = defteros = 0;
    queue<int> tmp;
    vector<int> Pascal;

    if (rowIndex == 0)
    {
        Pascal.push_back(1);
        return Pascal;
    }                         

    tmp.push(0);
    tmp.push(1);
    tmp.push(1);

    for (int i = 1; i < rowIndex; i++)
    {
        tmp.push(0);
        do
        {
            aspros = tmp.front();
            if (!tmp.empty())
                tmp.pop();
            defteros = tmp.front();
            tmp.push(aspros + defteros);
        } while (defteros != 0);
    }
    tmp.pop();
    while (!tmp.empty())
    {
        Pascal.push_back(tmp.front());
        tmp.pop();
    }

    return Pascal;
}

queue和vector其实感觉还不熟，应该可以拿vector直接来做的【以后改】

[leetcode] 2. Pascal's Triangle II